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babymother [125]

3 years ago

5

A steel wire has a cross sectional area of 1.5 x 10^-4 cm^2 and a length of 10.0 m. By how much will it stretch if a 25 kg weigh

t is hung from it? Please answer in millimeters.

1 answer:

larisa86 [58]3 years ago

4 0

<u>Answer:</u> The steel wire will stretch up to 1837.5 mm

<u>Explanation:</u>

We are given:

Mass = 25 kg

Length of wire = 10 m

Area of cross-section of wire = $1.5\times 10^{-4}cm^2=1.5\times 10^{-8}m^2$ (Conversion factor: $1cm^2=10^{-4}m^2$ )

To calculate the change in stretching, we use the equation:

$E=\frac{Fl}{A\Delta l}$

where,

E = young modulus of steel = $20\times 10^{10}Pa$

F = force exerted by the weight = m g = $25kg\times 9.8m/s^2$

l = length of wire = 10 m

A = area of cross section = $1.5\times 10^{-8}m^2$

$\Delta l$ = change in length = ?

Putting values in above equation, we get:

$20\times 10^{10}=\frac{25\times 9.8\times 10}{1.5\times 10^{-8}\times \Delta l}\\\\\Delta l=1.8375m$

Converting above value in mili meters, we use the conversion factor:

1 m = 1000 mm

So, 1.8375 m = 1837.5 mm

Hence, the steel wire will stretch up to 1837.5 mm

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Two parallel, circular conducting plates 32 cm in diameter are separated by 0.5 cm and have charges of +24 nC and -24 nC, respec

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Answer: E = 33762.39 N/c

Explanation: we calculate the capacitance of the two conducting plates ( this is because, 2 circular disc carrying opposite charges and the same dimension form a capacitor).

C = A/4πkd

Where C = capacitance of capacitor

A = Area of plates = πr² ( where r is radius which is half of the diameter)

K = electric constant = 9×10^9

d = distance between plates = 0.5cm = 0.005 m

Let us get the area, A = πr², where r = D/2 where D = diameter

r = 32/2 = 16cm = 0.16m

A = 22/7 × (0.16)² = 0.0804 m²

By substituting this into the capacitance formula, we have that

C = 0.0804/4×3.142*9×10^9 × 0.005

C = 0.0804/565486677.646

C = 142.17*10^(-12) F.

But C =Q/V where V = Ed

Hence we have that

C = Q/Ed

Where C = capacitance of capacitor = 142.17*10^(-12)F

Q = magnitude of charge on the capacitor = 24×10^-9c

E = strength of electric field =?

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142.17*10^(-12) = 24 ×10^-9 / E × 0.005

By cross multiplying

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8 0

3 years ago

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

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Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

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$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago

A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 2.3 cm from the axis of rotation. (a) Calcul

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Answer:

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$a_{r} = 0.275\,\frac{m}{s^{2}}$

b) The watermelon seed is experimenting a centrifugal acceleration. The coefficient of static friction between the seed and the turntable is calculated by the Newton's Laws:

$\Sigma F = \mu_{s}\cdot m\cdot g = m\cdot a$

$a = \mu_{s}\cdot g$

$\mu_{s} = \frac{a}{g}$

$\mu_{s} = \frac{0.275\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }$

$\mu_{s} = 0.028$

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$\alpha = \frac{\omega-\omega_{o}}{\Delta t}$

$\alpha = \frac{3.456\,\frac{rad}{s}-0\,\frac{rad}{s} }{0.36\,s}$

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$a = \sqrt{a_{t}^{2}+a_{r}^{2}}$

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4 0

3 years ago

A set of four capacitors are attached to a 12V battery in the circuit shown below. All capacitances are measured in milli-Farads

Bond [772]

The amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.

<h3>Total capacitance of the circuit</h3>

The total capacitance of the circuit is calculated as follows;

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Q = 0.03087 x 12

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Thus, the amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.

Learn more about capacitors here: brainly.com/question/13578522

#SPJ1

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